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C Exercises: Check whether an alphabet is a vowel or consonant

C Conditional Statement: Exercise-17 with Solution

Write a C program to check whether an alphabet is a vowel or consonant.

Pictorial Presentation:

Check whether an alphabet is a vowel or consonant

Sample Solution:

C Code:

#include <stdio.h>
void main()  
{  
    char sing_ch;  
    
    printf("Input any alphabet : ");  
    scanf("%c", &sing_ch);  
  
    if(sing_ch=='a' || sing_ch=='e' || sing_ch=='i' || sing_ch=='o' || sing_ch=='u' || sing_ch=='A' || sing_ch=='E' || sing_ch=='I' || sing_ch=='O' 

|| sing_ch=='U')  
    {  
        printf("The alphabet is a vowel.\n");  
    }  
    else if((sing_ch>='a' && sing_ch<='z') || (sing_ch>='A' && sing_ch<='Z'))  
    {  
        printf("The alphabet is a consonant.\n");  
    }  
    else  
    {  
        printf("The character is not an alphabet.\n");  
    }   
}
 

Sample Output:

Input any alphabet : K                                                                                        
The alphabet is a consonant. 

Flowchart:

Flowchart: Check whether an alphabet is vowel or consonant

C Programming Code Editor:

Improve this sample solution and post your code through Disqus.

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C Programming: Tips of the Day

Static variable inside of a function in C

The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().

The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.

The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().

Ref : https://bit.ly/3fOq7XP